Question

The point (1, −1) is on the terminal side of angle θ, in standard position. What are the values of sine, cosine, and tangent of θ? Make sure to show all work.

Answer 1

I have to request this early because I don't have a clue what the unit coordinate plane is and this is entirely new to me. Would appreciate a good expplainer

Answer 2

explainer*

Answer 3

I see

Answer 4

right one , down one

Answer 5

Right

Answer 6

you need only one other number, the length of this hypotenuse

Answer 7

you find that via pythagoras

Answer 8

What is pythagoras?

Answer 9

for a right triangle with legs \(a,b\) and hypotenuse \(h\) then \[a^2+b^2=h^2\]

Answer 10

in your case here, \(a\) is \(1\) and \(b\) is also \(1\) so it is just \[1^2+1^2=h^2\]

Answer 11

ok

Answer 12

so... what is \(h\)?

Answer 13

2

Answer 14

no, \(h^2=2\) you need \(h\)

Answer 15

ok

Answer 16

h^2=2

Answer 17

yeah so \(h=?\)

Answer 18

h=square root of 2

Answer 19

yes \[h=\sqrt2\]

Answer 20

cool

Answer 21

now you can find any trig ratio you like
cosine is the first coordinate over the hypotenuse
sine is the second coordinate over the hypotenuse
tangent is the second coordinate over the first coordinate

Answer 22

neato

Answer 23

i suppose
be advised that most people would not write \[\cos(\theta)=\frac{1}{\sqrt2}\] but rather \[\cos(\theta)=\frac{\sqrt2}{2}\] is is the same number, just with the denominator rationalized

Answer 24

? I dont seem to follow sorry

Answer 25

what, the last thing i wrote?

Answer 26

yes

Answer 27

you get that cosine is one over root two yes?

Answer 28

Like sohcahtoa right? I remember that

Answer 29

yes like that, only with coordinates

Answer 30

right

Answer 31

what i was trying to say was that instead of writing \[\frac{1}{\sqrt2}\] here for cosine, must people instead write \[\frac{\sqrt2}{2}\] i.e. they "rationalize the denominator"

Answer 32

oh ok

Answer 33

thanks again

Answer 34

and nice username